线性代数求助
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增广矩阵 (A, b) =
[ 1 -1 0 0 0 a1]
[ 0 1 -1 0 0 a2]
[ 0 0 1 -1 0 a3]
[ 0 0 0 1 -1 a4]
[-1 0 0 0 1 a5]
第 1,2,3,4 行加到第 5 行,初等变换为
[ 1 -1 0 0 0 a1]
[ 0 1 -1 0 0 a2]
[ 0 0 1 -1 0 a3]
[ 0 0 0 1 -1 a4]
[ 0 0 0 0 0 a1+a2+a3+a4+a5]
方程组有解的充要条件是 r(A, b) = r(A) , 则 a1+a2+a3+a4+a5 = 0.
此时方程组同解变形为
x1 - x2 = a1
x2 - x3 = a2
x3 - x4 = a3
x4 = a4 + x5
取 x5 = 0 , 得特解 (a1+a2+a3+a4, a2+a3+a4, a3+a4, a4, 0)^T
导出组是
x1 - x2 = 0
x2 - x3 = 0
x3 - x4 = 0
x4 = x5
取 x5 = 1 , 得基础解系 (1, 1, 1, 1, 1)^T
则通解是
x = (a1+a2+a3+a4, a2+a3+a4, a3+a4, a4, 0)^T + k(1, 1, 1, 1, 1)^T
[ 1 -1 0 0 0 a1]
[ 0 1 -1 0 0 a2]
[ 0 0 1 -1 0 a3]
[ 0 0 0 1 -1 a4]
[-1 0 0 0 1 a5]
第 1,2,3,4 行加到第 5 行,初等变换为
[ 1 -1 0 0 0 a1]
[ 0 1 -1 0 0 a2]
[ 0 0 1 -1 0 a3]
[ 0 0 0 1 -1 a4]
[ 0 0 0 0 0 a1+a2+a3+a4+a5]
方程组有解的充要条件是 r(A, b) = r(A) , 则 a1+a2+a3+a4+a5 = 0.
此时方程组同解变形为
x1 - x2 = a1
x2 - x3 = a2
x3 - x4 = a3
x4 = a4 + x5
取 x5 = 0 , 得特解 (a1+a2+a3+a4, a2+a3+a4, a3+a4, a4, 0)^T
导出组是
x1 - x2 = 0
x2 - x3 = 0
x3 - x4 = 0
x4 = x5
取 x5 = 1 , 得基础解系 (1, 1, 1, 1, 1)^T
则通解是
x = (a1+a2+a3+a4, a2+a3+a4, a3+a4, a4, 0)^T + k(1, 1, 1, 1, 1)^T
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